是否可以根据给定条件分别从1和0达到N和M

给定两个整数N和M ，任务是检查是否可以通过多次执行两个操作分别从X = 1和Y = 0获得这些值：

• 当且仅当x> 0时，将X和Y加1。
• 当且仅当y> 0时，将Y增加2。

例子：

Input: N = 3, M = 4
Output: Yes
Explanation: 
Initially X = 1, Y = 0 
Operation 1: X = 2, Y = 1
Operation 1: X = 3, Y = 2 
Operation 2: X = 3, Y = 4, hence the final values are got so the answer is Yes.

Input: N = 5, M = 2
Output: No 
Explanation : 
Obtaining X = 5 and Y = 2 from X = 1 and Y = 0 is not possible.

方法：可以使用以下观察结果解决以上问题：

1. 如果N小于2并且M不等于零，则不可能获得最终值，因此答案为No。
2. 否则，从M减去N，如果M? 0并且M被2整除则答案是是。
3. 在其他所有情况下，答案都是“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function that find given x and y
// is possible or not
bool is_possible(int x, int y)
{
    // Check if x is less than 2 and
    // y is not equal to 0
    if (x < 2 && y != 0)
        return false;
 
    // Perform subtraction
    y = y - x + 1;
 
    // Check if y is divisible by 2
    // and greater than equal to 0
    if (y % 2 == 0 && y >= 0)
        return true;
 
    else
        return false;
}
 
// Driver Code
int main()
{
    // Given X and Y
    int x = 5, y = 2;
 
    // Function Call
    if (is_possible(x, y))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program for the above approach
class GFG{
  
// Function that find given x and y
// is possible or not
static boolean is_possible(int x, int y)
{
    // Check if x is less than 2 and
    // y is not equal to 0
    if (x < 2 && y != 0)
        return false;
  
    // Perform subtraction
    y = y - x + 1;
  
    // Check if y is divisible by 2
    // and greater than equal to 0
    if (y % 2 == 0 && y >= 0)
        return true;
  
    else
        return false;
}
  
// Driver Code
public static void main(String[] args)
{
    // Given X and Y
    int x = 5, y = 2;
  
    // Function Call
    if (is_possible(x, y))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by rock_cool

Python3

# Python3 program for the above approach
 
# Function that find given x and y
# is possible or not
def is_possible(x, y):
   
    # Check if x is less than 2 and
    # y is not equal to 0
    if (x < 2 and y != 0):
        return false
 
    # Perform subtraction
    y = y - x + 1
 
    # Check if y is divisible by 2
    # and greater than equal to 0
    if (y % 2 == 0 and y >= 0):
        return True
    else:
        return False
 
# Driver Code
if __name__ == '__main__':
   
    # Given X and Y
    x = 5
    y = 2
 
    # Function Call
    if (is_possible(x, y)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by Mohit Kumar

C#

// C# program for the above approach
using System;
class GFG{
   
// Function that find given x and y
// is possible or not
static bool is_possible(int x, int y)
{
    // Check if x is less than 2 and
    // y is not equal to 0
    if (x < 2 && y != 0)
        return false;
   
    // Perform subtraction
    y = y - x + 1;
   
    // Check if y is divisible by 2
    // and greater than equal to 0
    if (y % 2 == 0 && y >= 0)
        return true;
   
    else
        return false;
}
   
// Driver Code
public static void Main(string[] args)
{
    // Given X and Y
    int x = 5, y = 2;
   
    // Function Call
    if (is_possible(x, y))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Ritik Bansal

Javascript

 
// Javascript program for the above approach
 
// Function that find given x and y
// is possible or not
function is_possible(x, y)
{
     
    // Check if x is less than 2 and
    // y is not equal to 0
    if (x < 2 && y != 0)
        return false;
 
    // Perform subtraction
    y = y - x + 1;
 
    // Check if y is divisible by 2
    // and greater than equal to 0
    if (y % 2 == 0 && y >= 0)
        return true;
    else
        return false;
}
 
// Driver code
 
// Given X and Y
let x = 5, y = 2;
 
// Function Call
if (is_possible(x, y))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by divyesh072019       
         

输出：

No

时间复杂度： O(1)
辅助空间： O(1)